Re: [9fans] portability question

[hugo rivera <uair00@gmail.com>]

Thanks for the feedback.

2010/6/16 Bakul Shah <bakul+plan9@bitblocks.com>:
> On Wed, 16 Jun 2010 11:11:09 +0200 hugo rivera <uair00@gmail.com>  wrote:
>> Can someone clarify why the program included outputs 'AB000000' (as I
>> expect) on 32 bit systems and 'FFFFFFFFAB000000' on 64 bit systems?
>> where all those 1's came from? what's the portable way of doing this?
>> sorry for newbie questions like this.
>>
>>
>>        unsigned long l;
>>         unsigned char c;
>>
>>         l = 0L;
>>         c = 0xAB;
>>         l |= c << 24;
>>         printf("%lX\n", l);
>
> For use of C on non-plan9 machines I would recommend
> downloading the last draft of the C9x standard as a ready
> reference.  Google for n843.
>
> As per Section 6.5.7 of C9x, both operands of << must be of
> "integer type" and the result type is that of the left
> operand. Since sizeof c < sizeof(int), it is promoted.  Now
> 6.3.1.2 says "if an int can represent all values of the
> original type, the value is converted to an int". So c is
> first converted to an int which means c << 24 is an integer
> and -ve. Since an int is smaller than a long (in your case)
> it is promoted to a long. Changing the |= statement to
>
>         l |= (unsigned)c << 24;
>
> should give you what you want.
>
>



-- 
Hugo